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At the bottom of the wikipedia article on climate proxies there is this formula: $$ \delta \, {\rm ^{18}O} = \frac{\frac{\left[\rm{^{18}O}\right]}{\left[\rm{^{16}O}\right]}}{\frac{\left[\rm{^{18}O}\right]}{\left[\rm{^{16}O}\right]} \Bigg|_{\rm VSMOW}} - 1 $$ and similar for $\delta \, {\rm D}$.

I'm wondering about this sentence: "$\delta$ values for precipitation are always negative." Why is that? In fact there is this figure at IPCC AR4 § 6.4, depicted below, that shows both $\delta$ values and to me it seems that indeed ${\rm \delta D}$ is always negative around $-42\%$, but ${\rm \delta ^{18} O}$ looks positive with values around $0.4\%$.

Figure 6-3 from IPCC AR4, which shows several paleoclimatic proxies, including $\delta \text D$ and $\delta^{18}\text O$.

Ok, how do these values arise? I tried to look at Raoult's law and estimated the ${\rm D_2O/H_2O}$ ratio (maybe it is even ${\rm HDO}$ instead of ${\rm D_2O}$, but it doesn't change the essence I guess) by $$ \frac{\left[\rm{D_2O}\right]}{\left[\rm{H_2O}\right]} \Bigg|_{\rm VSMOW} = \frac{x_{\rm D_2O} \, p_{\rm D_2O}}{p_{\rm H_2O}} $$ where $x$ is the mixing ratio of ${\rm D_2O}$ in the sea-water and $p$ is the saturation vapour pressure. I gathered up some values from Besley, L. and Bottomley, G.A., 1973, Vapour pressure of normal and heavy water from 273.15 to 298.15 K, The Journal of Chemical Thermodynamics and looked at the temperature dependence.

Plot of ratio of partial pressure of heavy water ($\text D_2\text O$) to that of normal water ($\text H_2\text O$) as a function of temperature.

Assuming for simplicity we start at ${\rm 25°C}$ when the water evaporates over the sea, then when it reaches the artic region where it condenses the ratio $p_{\rm D_2O}/p_{\rm H_2O}$ will have changed and as I assume the equilibrium concentration ratio will change too. From the above plot (even though it only goes down to ${\rm 4°C}$) we see that the ratio drops, so I pressume more ${\rm D_2O}$ condenses compared to ${\rm H_2O}$ which increases the ratio $$ \frac{\left[\rm{D_2O}\right]}{\left[\rm{H_2O}\right]} $$ found in ice-core samples, but that does not agree with the statement that for precipitation values are always negative.

Does anyone know more?

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I'm wondering about this sentence: "$\delta$ values for precipitation are always negative." Why is that?

Always? That is far too strong, and the referenced citation does not support that claim. However, a weaselly word such as "usually" would be far too weak. Except for the overly-strong "always", there is a big grain of truth to that claim.

As you noted later in your question, D2O precipitates more readily than does H2O. The same applies to semi-heavy water (DHO) and to H218O. You are missing the flip side of this, which is that those heavier forms of water evaporate less readily than does H2O. This, I think, forms the core of your misunderstanding.

Precipitation comes from water vapor in the atmosphere, which in turn comes from water evaporated from surface water. Evaporated water is always depleted of deuterium and 18O compared to the water from which the vapor evaporated. Water that condenses in the air always has enhanced levels of deuterium and 18O compared to the vapor from which it condensed. However, the water that eventually falls as rain condensed from 18O-depleted water vapor. This means that - Precipitation, even near the equator, is usually (not always) depleted of deuterium and 18O. - The vapor that doesn't fall as precipitation is even more strongly depleted of deuterium and 18O than it was before some of that vapor condensed and fell as precipitation.

Some of this vapor migrates toward the poles, and each step along the way gets ever more depleted of the heavier isotopes of hydrogen and oxygen. The water vapor at high latitudes, and hence the precipitation at such latitudes always is depleted of those heavier isotopes.

... but $\delta^{18}O$ looks positive with values around 0.4%.

You misread the referenced figure, reproduced below.

Figure 6.3 from IPCC AR4 Chapter 6, which shows several proxies for temperature, including $\delta D$ in the Vostok ice core and $\delta^{18}O$ in benthic (deep sea) sediment samples.

The $\delta^{18}O$ levels are indeed enhanced in that figure, but you need to look at what this graph depicts. It shows benthic (i.e., deep sea) $\delta^{18}O$ levels as measured from cores taken from deep sea sediments. Since precipitation in high latitudes is depleted of heavier isotopes of hydrogen and oxygen, it stands to reason that waters left behind in equatorial regions are enhanced in those heavier isotopes. The enhancement increases when ice sheets build up, but then decreases thousands of years later when the ice eventually melts.

The $\delta D$ plot in the above graph result from an ice core taken from the Vostok station in Antarctica. The study that provided the $\delta D$ plot in the above graph also showed $\delta^{18}O$ levels in that ice core. The $\delta^{18}O$ curve showed the same general characteristics as the $\delta D$ curve but was much coarser. The cores taken from deep sea sediments as expected showed the reverse behavior, but with a much finer resolution. With regard to $\delta^{18}O$ levels, those benthic samples make for a better temperature proxy than do ice cores.

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  • $\begingroup$ 1. So deep sea sediments supply good preservation for samples over millions of years similar to the effect of ice cores? You state: Since precipitation in high latitudes is depleted of heavier isotopes of hydrogen and oxygen, it stands to reason that waters left behind in equatorial regions are enhanced in those heavier isotopes. The enhancement increases when ice sheets build up, but then decreases thousands of years later when the ice eventually melts. But then ice only forms at the higher latitude and how are you then talking about ice sheets at the equator? $\endgroup$ – Diger Aug 2 '18 at 10:00
  • $\begingroup$ 2. You are missing the flip side of this, which is that those heavier forms of water evaporate less readily than does ${\rm H_2O}$. This, I think, forms the core of your misunderstanding. Isn't that what I state with Raoult's law above? So one point is that while air travels to higher latitudes, heavier isotopes are depleted from the air, but on the flip-side when it condenses at the poles the condensed water always has a higher isotopic ratio than the vapour it condenses from. Doesn't this mean that these two effects cancel partially? $\endgroup$ – Diger Aug 2 '18 at 10:02
  • $\begingroup$ @Diger - Re how are you then talking about ice sheets at the equator? I'm not. You are missing a key aspect of the hydrological cycle, which is that the atmosphere transports water from poleward from the equator and the tropics. Re your second comment, Doesn't this mean that these two effects cancel partially?, yes, they do, but the precipitant is always depleted in heavy isotopes compared to the liquid water from which the precipitant arose because of the heavy isotope depletion in the vapor from which the precipitant condensed. $\endgroup$ – David Hammen Aug 2 '18 at 13:37
  • $\begingroup$ yes, they do, but the precipitant is always depleted in heavy isotopes compared to the liquid water from which the precipitant arose because of the heavy isotope depletion in the vapor from which the precipitant condensed I'm sorry, but I do not get at all, what you try to tell me :-( What do you mean? The precipitant is depleted preferentially compared to the liquid water from which the precipitant arose? What about the last part depletion in the vapor from which the precipitant condensed ? $\endgroup$ – Diger Aug 2 '18 at 14:06
  • $\begingroup$ @Diger - First off, I shouldn't have written "always". I can imagine some rare circumstances under which the precipitant is enhanced in heavy isotopes compared to the liquid water from which the precipitant originally arose. However, that's got to be rather rare. $\endgroup$ – David Hammen Aug 2 '18 at 14:28

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